Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n + 7}{-3n^2 - 3n + 216} \times \dfrac{n^3 + 3n^2 - 54n}{n^2 - 6n} $
Solution: First factor out any common factors. $a = \dfrac{n + 7}{-3(n^2 + n - 72)} \times \dfrac{n(n^2 + 3n - 54)}{n(n - 6)} $ Then factor the quadratic expressions. $a = \dfrac {n + 7} {-3(n + 9)(n - 8)} \times \dfrac {n(n + 9)(n - 6)} {n(n - 6)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(n + 7) \times n(n + 9)(n - 6) } { -3(n + 9)(n - 8) \times n(n - 6)} $ $a = \dfrac {n(n + 9)(n - 6)(n + 7)} {-3n(n + 9)(n - 8)(n - 6)} $ Notice that $(n + 9)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {n\cancel{(n + 9)}(n - 6)(n + 7)} {-3n\cancel{(n + 9)}(n - 8)(n - 6)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $a = \dfrac {n\cancel{(n + 9)}\cancel{(n - 6)}(n + 7)} {-3n\cancel{(n + 9)}(n - 8)\cancel{(n - 6)}} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac {n(n + 7)} {-3n(n - 8)} $ $ a = \dfrac{-(n + 7)}{3(n - 8)}; n \neq -9; n \neq 6 $